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# Numbers formed by flipping common set bits in two given integers

• Difficulty Level : Medium
• Last Updated : 08 Apr, 2021

Given two positive integers A and B, the task is to flip the common set bitsin A and B.

Examples:

Input: A = 7, B = 4
Output: 3 0
Explanation:
The binary representation of 7 is 111
The binary representation of 4 is 100
Since the 3rd bit of both A and B is a set bit. Therefore, flipping the 3rd bit of A and B modifies A = 3 and B = 0
Therefore, the required output is 3 0

Input: A = 10, B = 20
Output: 10 20

Naive Approach: The simplest approach to solve this problem is to check if the ith bit of both A and B is a set or not. If found to be true, then flip the ith bit of both A and B. Finally, print the updated values of both A and B.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to flip bits of A and B``// which are set bits in A and B``void` `flipBitsOfAandB(``int` `A, ``int` `B)``{``    ``// Iterater all possible bits``    ``// of A and B``    ``for` `(``int` `i = 0; i < 32; i++) {` `        ``// If ith bit is set in``        ``// both A and B``        ``if` `((A & (1 << i)) && (B & (1 << i))) {` `            ``// Clear i-th bit of A``            ``A = A ^ (1 << i);` `            ``// Clear i-th bit of B``            ``B = B ^ (1 << i);``        ``}``    ``}` `    ``// Print A and B``    ``cout << A << ``" "` `<< B;``}` `// Driver Code``int` `main()``{``    ``int` `A = 7, B = 4;` `    ``flipBitsOfAandB(A, B);` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;``  ` `class` `GFG{``  ` `// Function to flip bits of A and B``// which are set bits in A and B``static` `void` `flipBitsOfAandB(``int` `A, ``int` `B)``{``    ` `    ``// Iterater all possible bits``    ``// of A and B``    ``for``(``int` `i = ``0``; i < ``32``; i++)``    ``{``        ` `        ``// If ith bit is set in``        ``// both A and B``        ``if` `(((A & (``1` `<< i)) &``             ``(B & (``1` `<< i))) != ``0``)``        ``{``            ` `            ``// Clear i-th bit of A``            ``A = A ^ (``1` `<< i);`` ` `            ``// Clear i-th bit of B``            ``B = B ^ (``1` `<< i);``        ``}``    ``}`` ` `    ``// Print A and B``    ``System.out.print(A + ``" "` `+ B);``}``  ` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `A = ``7``, B = ``4``;`` ` `    ``flipBitsOfAandB(A, B);``}``}` `// This code is contributed by code_hunt`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to flip bits of A and B``# which are set in both of them``def` `flipBitsOfAandB(A, B):``    ` `    ``# Iterate all possible bits of``    ``# A and B``    ``for` `i ``in` `range``(``0``, ``32``):``        ` `        ``# If ith bit is set in``        ``# both A and B``        ``if` `((A & (``1` `<< i)) ``and``            ``(B & (``1` `<< i))):``            ` `            ``# Clear i-th bit of A``            ``A ``=` `A ^ (``1` `<< i)` `            ``# Clear i-th bit of B``            ``B ``=` `B ^ (``1` `<< i)``            ` `    ``print``(A, B)` `# Driver Code ``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``A ``=` `7``    ``B ``=` `4``    ` `    ``flipBitsOfAandB(A, B)``    ` `# This code is contributed by Virusbuddah_`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{` `// Function to flip bits of A and B``// which are set bits in A and B``static` `void` `flipBitsOfAandB(``int` `A, ``int` `B)``{``    ` `    ``// Iterater all possible bits``    ``// of A and B``    ``for``(``int` `i = 0; i < 32; i++)``    ``{``        ` `        ``// If ith bit is set in``        ``// both A and B``        ``if` `(((A & (1 << i)) &``             ``(B & (1 << i))) != 0)``        ``{``            ` `            ``// Clear i-th bit of A``            ``A = A ^ (1 << i);` `            ``// Clear i-th bit of B``            ``B = B ^ (1 << i);``        ``}``    ``}` `    ``// Print A and B``    ``Console.Write(A + ``" "` `+ B);``}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``    ``int` `A = 7, B = 4;` `    ``flipBitsOfAandB(A, B);``}``}` `// This code is contributed by chitranayal`

## Javascript

 ``
Output:
`3 0`

Time Complexity: O(32)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is based on the following observations:

Find the bits which are set bits in both A and B using (A & B).
Clear the bits of A which are set bits in both A and B using A = (A ^ (A & B))
Cleat the bits of B which are set bits in both A and B using B = (B ^ (A & B))

Follow the steps below to solve the problem:

• Update A = (A ^ (A & B)).
• Update B = (B ^ (A & B)).
• Finally, print the updated values of A and B.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to flip bits of A and B``// which are set in both of them``void` `flipBitsOfAandB(``int` `A, ``int` `B)``{` `    ``// Clear the bits of A which``    ``// are set in both A and B``    ``A = A ^ (A & B);` `    ``// Clear the bits of B which``    ``// are set in both A and B``    ``B = B ^ (A & B);` `    ``// Print updated A and B``    ``cout << A << ``" "` `<< B;``}` `// Driver Code``int` `main()``{``    ``int` `A = 10, B = 20;` `    ``flipBitsOfAandB(A, B);` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;``   ` `class` `GFG{``   ` `// Function to flip bits of A and B``// which are set in both of them``static` `void` `flipBitsOfAandB(``int` `A, ``int` `B)``{``    ` `    ``// Clear the bits of A which``    ``// are set in both A and B``    ``A = A ^ (A & B);``    ` `    ``// Clear the bits of B which``    ``// are set in both A and B``    ``B = B ^ (A & B);`` ` `    ``// Print updated A and B``    ``System.out.print(A + ``" "` `+ B);``}``   ` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `A = ``10``, B = ``20``;``    ` `    ``flipBitsOfAandB(A, B);``}``}` `// This code is contributed by sanjoy_62`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to flip bits of A and B``# which are set in both of them``def` `flipBitsOfAandB(A, B):``    ` `    ``# Clear the bits of A which``    ``# are set in both A and B``    ``A ``=` `A ^ (A & B)` `    ``# Clear the bits of B which``    ``# are set in both A and B``    ``B ``=` `B ^ (A & B)` `    ``# Print updated A and B``    ``print``(A, B)` `# Driver Code ``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``A ``=` `10``    ``B ``=` `20``    ` `    ``flipBitsOfAandB(A, B)``    ` `# This code is contributed by Virusbuddah_`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{``   ` `// Function to flip bits of A and B``// which are set in both of them``static` `void` `flipBitsOfAandB(``int` `A, ``int` `B)``{``  ` `    ``// Clear the bits of A which``    ``// are set in both A and B``    ``A = A ^ (A & B);``    ` `    ``// Clear the bits of B which``    ``// are set in both A and B``    ``B = B ^ (A & B);`` ` `    ``// Print updated A and B``    ``Console.Write(A + ``" "` `+ B);``}``   ` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `A = 10, B = 20;``    ` `    ``flipBitsOfAandB(A, B);``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``
Output:
`10 20`

Time Complexity: O(1)
Auxiliary Space: O(1)

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